Question

A website advertises job openings on its website, but job seekers have to pay to access the list of job openings. The website recently completed a survey to estimate the number of days it takes to find a new job using its service. It took the last 30 customers an average of 60 days to find a job. Assume the population standard deviation is 10 days. Construct a 90% confidence interval of the population mean number of days it takes to find a job.

Answer 1

Answer:

The 90% confidence interval would be given by (57.006;62.994)  

Step-by-step explanation:

Previous concepts

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

$\bar X=60$ represent the sample mean  

$\mu$ population mean (variable of interest)  

$\sigma=10$ represent the population standard deviation  

n=30 represent the sample size  

Assuming the X follows a normal distribution  

$X \sim N(\mu, \sigma=10)$

The sample mean $\bar X$ is distributed on this way:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$  

The confidence interval on this case is given by:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

The next step would be find the value of $\z_{\alpha/2}$, $\alpha=1-0.90=0.1$,$\alpha/2 =0.05$ and $z_\alpha/2=1.64$  

Using the normal standard table, excel or a calculator we see that:  

$z_{\alpha/2}=1.64$

Since we have all the values we can replace:

$60 - 1.64\frac{10}{\sqrt{30}}=57.006$  

$60 + 1.64\frac{10}{\sqrt{30}}=62.994$  

So on this case the 90% confidence interval would be given by (57.006;62.994)